Arithmetic evaluation: Difference between revisions

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@@ Line 4,520: / Line 4,520: @@
 procedure PopFactor()
-object p2 = opstack[$]
+    object p1, p2 = opstack[$]
     if op="u-" then
+        p1 = 0
-        if op_p_p=1 then                        -- op_p_p
-            opstack[$] = {op,0,p2}
-        elsif op_p_p=0 then                     -- p_op_p
-            opstack[$] = {0,op,p2}
-        else -- -1                              -- p_p_op
-            opstack[$] = {0,p2,op}
-        end if
     else
         opstack = opstack[1..$-1]
+        p1 = opstack[$]
-        if op_p_p=1 then                        -- op_p_p
+    end if
-            opstack[$] = {op,opstack[$],p2}
-        elsif op_p_p=0 then                     -- p_op_p
+    if op_p_p=1 then
-            opstack[$] = {opstack[$],op,p2}
+        opstack[$] = {op,p1,p2}  -- op_p_p
+    elsif op_p_p=0 then
-        else -- -1                              -- p_p_op
-            opstack[$] = {opstack[$],p2,op}
+        opstack[$] = {p1,op,p2}  -- p_op_p
-        end if
+    else -- -1
+        opstack[$] = {p1,p2,op}  -- p_p_op
     end if
     op = 0
@@ Line 4,665: / Line 4,660: @@
 I added a flag (for this task) to store the ast nodes as op_p_p, p_op_p, or p_p_op, whichever you prefer.
 {{out}}
+For "3+4+5+6*7/1*5^2^3", the fully parenthesised Phix equivalent being ((3+4)+5)+(((6*7)/1)*power(5,power(2,3)))
 <pre>
 with op_p_p: